$$\int\frac{-1}{3(2x-1)}dx$$ there are two answers. One is $$\frac{-1}{6}\ln(2x-1)+C$$, if you take the $\frac{-1}{3}$ outside the integral immediately. u is 2x-1. However, if you multiply the 3 to the 2x-1, making it $$\int\frac{-1}{(6x-3)}dx$$, u becomes 6x-3, and the answer is now $$\frac{-1}{6}\ln(6x-3)+C$$ what is going on here??????
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2$\ln(6x-3)=\ln(2x-1)+\text{ constant }$ – J. W. Tanner Jun 05 '19 at 02:10
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Hint: $\ln(ab)=\ln(a)+\ln(b)$ – randomgirl Jun 05 '19 at 02:10
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Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. – dantopa Jun 05 '19 at 02:13
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Some the best exercises in elementary mathematics arise in integration when trying to show that two seemingly different answers are in fact the same. For instance, if yo try to integrate $\sin x \cos x$ with respect to $x.$ What happens if you let $u=\sin x?$ Repeat with $u = \cos x.$ The two answers are equal up to a change in constant provided by the fundamental trigonometric identity. In time, you learn to watch for these situations, as with the fact about logarithms in the previous comments and the answers below. – Chris Leary Jun 05 '19 at 02:33
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2Possible duplicate of Two different solutions to integral – Hans Lundmark Jun 05 '19 at 16:40
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Nothing wrong, $\ln(AB)=\ln(A)+\ln(B), \ln(6x-3)=\ln(3)+\ln(2x-1)$, so $\ln(2x-1)+C=\ln(6x-3)+C'$ where $C=\ln(3)+C'$.
xrfxlp
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Tsemo Aristide
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OneIndefinite integral $I(x)$ does by two methods may give two different looking expressions say $I_1(x)$ and $I_2(x)$ and their difference is akways a contant (independent of $x$). Check that in your case both expressions differ by $\pm \frac{1}{6} \ln 3.$
More interesting example is off $I=\int \sin x \cos x dx= \frac{1}{2}\int \sin 2x dx=-\frac{\cos 2x}{4}+C =I_1(x)$, but if you use $\sin x =t$, you get $I_2(x)=\frac {\sin^2 x}{2}+D$. Notice that $I_1(x)-I_2(x)=C-D-\frac{1}{4}$ (a constant).
Z Ahmed
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Your two answers were $\displaystyle -\frac{1}{6}\ln(2x-1)+C_1$ and $\displaystyle -\frac{1}{6}\ln(6x-3)+C_2$, where $C$ is an arbitrary constant.
Because of the property that $\ln(A)+\ln(B)=\ln(AB)$, we see that $\displaystyle -\frac{1}{6}\ln(2x-1)-6\ln(3)=-\frac{1}{6}\ln(6x-3)$, and so both methods actually produce the same answer!
It's like how $\cos(x)$ and $\displaystyle \sin\left(x+\frac{π}{2}\right)$ are actually the same.
Saketh Malyala
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I think both answers are wrong!
The right answer it's: $$\int\frac{-1}{3(2x-1)}dx=-\frac{1}{6}\ln(2x-1)+C_1,$$ where $x>\frac{1}{2}$ and $$\int\frac{-1}{3(2x-1))}dx=-\frac{1}{6}\ln(1-2x)+C_2,$$ where $x<\frac{1}{2}.$
Michael Rozenberg
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